# Find Tangent Slope with Polar coordinates

1. May 9, 2010

### JSGhost

1. The problem statement, all variables and given/known data
Find the slope of the tangent line to the given polar curve at the point specified by the value of θ.
r = 9sin(θ)
θ = pi/6

2. Relevant equations
dy/dx = (dy/dθ) / (dx/dθ)
x=rcosθ
y=rsinθ

(sinx)^2 = (1/2)(1-cos2x)
(cosx)^2 = (1/2)(1+cos2x)
2sinxcosx = sin(2x)

3. The attempt at a solution
r = 9sin(θ)
x=rcos(θ)=9sin(θ)cos(θ)= --> Would this be sin(9θ) or sin(18θ)?
y=rsin(θ)=9sin(θ)sin(θ)= 9sin^2(θ)

dy/dx = (dy/dθ) / (dx/dθ) = (9 * 9sinθcosθ) / (cos(18θ) * 9) = sin(18θ)/cos(18θ) = tan(18θ)

θ = pi/6
tan(18*pi/6)=tan(3pi)=0

2. May 9, 2010

### tiny-tim

Welcome to PF!

Hi JSGhost! Welcome to PF!

(have a pi: π and try using the X2 tag just above the Reply box )
Nooo! you can't move numbers in and out of a function like sin (you wouldn't do it for √, would you? )
But you're not differentiating.

Start again.

3. May 9, 2010

### JSGhost

Thanks for replying. I actually figured it out after posting this topic. Saw a similar problem to this one on another site. Thanks anyways.

Find the slope of the tangent line to the given polar curve at the point specified by the value of θ.
r = 9sin(θ)
θ = pi/6

x=rcos(θ)=9sin(θ)cos(θ)= 9(sin(2θ))
y=rsin(θ)=9sin(θ)sin(θ)= 9sin^2(θ)

dx/dθ = 9(cos^2(θ) - sin^2(θ))
dy/dθ = 9(2sinθcosθ)

dy/dx = (dy/dθ)/(dx/dθ)
= 9(2sinθcosθ) / 9(cos^2(θ) - sin^2(θ))
= [2sinθcosθ] / [cos^2(θ)-sin^2(θ)]

when θ = pi/6
dy/dx = [2sin(pi/6)cos(pi/6)] / [cos^2(pi/6) - sin^2(pi/6)]
= [2*(1/2)*(sqrt(3)/2)] / [(sqrt(3)/2)*(sqrt(3)/2)-(1/2)*(1/2)]
= (sqrt(3)/2) / 2/4 = (sqrt(3)/2) / (1/2) = sqrt(3)